Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int n, m;
long long int ar[101][101], ar1[101][101];
vector<long long int> row, col, row1, col1;
inline long long int rc() {
  long long int maxi = -1, mini = 1e9;
  for (long long int i = 0; i < n; i++) {
    mini = 1e9;
    for (long long int j = 0; j < m; j++) {
      mini = min(mini, ar[i][j]);
    }
    for (long long int j = 0; j < m; j++) {
      ar[i][j] -= mini;
    }
    for (long long int j = 0; j < mini; j++) row.push_back(i + 1);
  }
  for (;;) {
    long long int x, y;
    maxi = -1;
    mini = 1e9;
    for (long long int i = 0; i < n; i++) {
      for (long long int j = 0; j < m; j++) {
        if (ar[i][j] > maxi) {
          maxi = ar[i][j];
          x = i;
          y = j;
        }
        mini = min(mini, ar[i][j]);
      }
    }
    if (mini < 0) {
      return 1e18;
    }
    if (maxi == 0) break;
    long long int cnth = 0, cntv = 0;
    for (long long int i = 0; i < m; i++) {
      if (ar[x][i] == 0) cnth++;
    }
    for (long long int i = 0; i < n; i++) {
      if (ar[i][y] == 0) cntv++;
    }
    if (cnth == 0) {
      row.push_back(x + 1);
      for (long long int i = 0; i < m; i++) ar[x][i]--;
    } else if (cntv == 0) {
      col.push_back(y + 1);
      for (long long int i = 0; i < n; i++) ar[i][y]--;
    } else {
      return 1e18;
    }
  }
  return row.size() + col.size();
}
inline long long int cr() {
  long long int maxi = -1, mini = 1e9;
  for (long long int i = 0; i < m; i++) {
    mini = 1e9;
    for (long long int j = 0; j < n; j++) {
      mini = min(mini, ar[j][i]);
    }
    for (long long int j = 0; j < n; j++) {
      ar[j][i] -= mini;
    }
    for (long long int j = 0; j < mini; j++) col1.push_back(i + 1);
  }
  for (;;) {
    long long int x, y;
    maxi = -1;
    mini = 1e9;
    for (long long int i = 0; i < n; i++) {
      for (long long int j = 0; j < m; j++) {
        if (ar[i][j] > maxi) {
          maxi = ar[i][j];
          x = i;
          y = j;
        }
        mini = min(mini, ar[i][j]);
      }
    }
    if (mini < 0) {
      return 1e18;
    }
    if (maxi == 0) break;
    long long int cnth = 0, cntv = 0;
    for (long long int i = 0; i < m; i++) {
      if (ar[x][i] == 0) cnth++;
    }
    for (long long int i = 0; i < n; i++) {
      if (ar[i][y] == 0) cntv++;
    }
    if (cntv == 0) {
      col1.push_back(y + 1);
      for (long long int i = 0; i < n; i++) ar[i][y]--;
    } else if (cnth == 0) {
      row1.push_back(x + 1);
      for (long long int i = 0; i < m; i++) ar[x][i]--;
    } else {
      return 1e18;
    }
  }
  return row1.size() + col1.size();
}
int main() {
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      cin >> ar[i][j];
      ar1[i][j] = ar[i][j];
    }
  }
  long long int RC = rc();
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      ar[i][j] = ar1[i][j];
    }
  }
  long long int CR = cr();
  long long int ans = min(RC, CR);
  if (ans == 1e18) {
    cout << -1;
    exit(0);
  }
  if (ans == RC) {
    cout << row.size() + col.size() << endl;
    for (long long int i = 0; i < row.size(); i++)
      cout << "row " << row[i] << endl;
    for (long long int i = 0; i < col.size(); i++)
      cout << "col " << col[i] << endl;
  } else {
    cout << row1.size() + col1.size() << endl;
    for (long long int i = 0; i < row1.size(); i++)
      cout << "row " << row1[i] << endl;
    for (long long int i = 0; i < col1.size(); i++)
      cout << "col " << col1[i] << endl;
  }
  return 0;
}